Let P(a_1,b_1) and Q(a_2,b_2) be two distinct points on a circle with centre C(√2,√3).

Question

Let $P(a_1,b_1)$ and $Q(a_2,b_2)$ be two distinct points on a circle with centre $C(\sqrt2,\sqrt3)$. Let $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of $	riangle OCP$ is $\dfrac{\sqrt{35}}2$, then $a_1^2+a_2^2+b_1^2+b_2^2$ is equal to \_\_\_\_\_\_\_.

Answer: 24

Accepted Answer

$OC=\sqrt{2+3}=\sqrt5$. Since $OC\perp CP$, $\text{Area}(\triangle OCP)=\tfrac12\cdot OC\cdot CP=\dfrac{\sqrt{35}}2\Rightarrow CP=\sqrt7$.

Since $P,Q$ lie on the circle centred at $C$, $CQ=CP=\sqrt7$ (both equal the radius). As $OC\perp CP$ and $OC\perp CQ$, triangles $OCP,OCQ$ are right-angled at $C$:
$$OP^2=OC^2+CP^2=5+7=12,\qquad OQ^2=5+7=12.$$

$a_1^2+b_1^2+a_2^2+b_2^2=OP^2+OQ^2=24.$

Solutions · 1